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Time of a falling object when the force of gravity isn't constant

  • #76
19,802
4,048
$$ t \simeq 2570 $$

I fixed posts #70 and #73, and now the results are the same, but still not sure since you got 2528.
That's because he used an inaccurate value of 0.78 for π/4. Your answer of 2570 is spot on.

Chet
 
  • #77
43
2
$$ t \simeq 2570 $$

I fixed posts #70 and #73, and now the results are the same, but still not sure since you got 2528.

$$\frac{{r_0}^{3/2}}{\sqrt{2 G M}} = \frac{{(2e7)}^{3/2}}{\sqrt{2 \times {6.67384e{-11}} \times {1.5e25}}} \simeq 1999$$

For this one, the drag related component of acceleration is related to speed^2, not position, so not quite the same. For a vertical drop, the problem can be solved, for the more general case of an object that also has a horizontal component of velocity, you need to use a numerical integration method.

As mentioned before a simpler problem for acceleration based on position is:
acceleration = -x
at time t = 0, x = 0, v = 1 meter / second
You'll be able to solve this and determine x(t), v(t) and a(t).
I learned alot from this question and I think,
with a bit more calculus I will be able to find the motion equations with air resistance
 
  • #78
rcgldr
Homework Helper
8,671
506
That's because he used an inaccurate value of 0.78 for π/4. Your answer of 2570 is spot on.
I cleaned this up a bit:

Given g = 10 m / s^2 at r = 1e7 m:
g = M G / r^2
G M = r^2 g = (1e7)^2 x 10 = 1e15
2 G M = 2e15
$$\frac{{r_0}^{3/2}}{\sqrt{2 G M}} = \frac{(2e7)^{3/2}}{\sqrt{2e15}} = 2e7 \sqrt{\frac{2e7}{2e15}} = 2e7 \sqrt{\frac{1}{2e8}} = \frac{2e7}{2e4} = 2000$$
$$t = -2000 ((π/4 - 1/2) -\ (π/2 - 0)) = 2000 (π/4 + 1/2) \simeq 2570.7963267949 $$

u = sqrt(r/r0-r) (versus r = r0 sin^2(θ))
This way is much clearer.
I'd recommend going back and looking at the two methods again. Although u = sqrt(r/r0-r) is a typical substitution method, in this case, with a bit of insight, using r = r0 sin^2(θ) replaced the square root component with tan(θ), and integrating 2 sin^2(θ)dθ is simpler than integrating (2 u^2 du) / (1 + u^2)^2 .
 
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  • #79
694
114
So is there no way to express ##r## in terms of ##t## only?
 
  • #80
19,802
4,048
So is there no way to express ##r## in terms of ##t## only?
Of course there is. Go to it rcgldr.

Chet
 
  • #81
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114
Btw, does this method assume that the planet remains stationary, or is the motion of both bodies included?
 
  • #82
694
114
I think I've found a neater solution to the problem. If you leave the equation in the form $$t=\int_{r_0}^r \frac{-dr}{\sqrt{\frac{2GM}{r} - \frac{2GM}{r_0}}}$$
and make the substitution ##r=\frac{r_0}{sec^2 \theta}##, you get
$$t=r_0 \sqrt{ \frac{r_0}{2GM}} (arccos(\sqrt{\frac{r}{r_0}})+\frac{\sqrt{r} \sqrt{r_0 -r}}{r_0})$$ with no pi or extra -ve signs.
(You can also arrive at this solution by considering the identity ##sin(\frac{\pi}{2} - \theta) = cos(\theta)## . Since the roles of the domain and range are switched with the inverse function, you get the identity ##\frac{\pi}{2} - arcsin(\theta) = arccos(\theta)## and arrive at the solution above.)

And so far, the only progress I've been able to make in using the function ##t(r)## to obtain the function ##r(t)## is the expression the solution in terms of a(n affine?) parameter ##\lambda## :
$$\sqrt{\frac{r}{r_0}}=cos \ \lambda ,$$ and
$$\frac{1}{r_0} \sqrt{\frac{2GM}{r_0}} t = cos \ \lambda (1+ sin \ \lambda)$$
I was thinking about using these Taylor sequences to go further:
$$cos \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n}}{(2n)!}$$
$$sin \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n+1}}{(2n+1)!}$$
However, this would then mean that the functions ##r(t)##, ##v(t)## and ##a(t)## would all have an infinite series, which would look very odd (pardon the parity pun). Is there any other way around this?

On a side note, doesn't the equation ##\frac{dr}{dt}=-\frac{dx}{dt}## assume that ##M>>m## so that ##M## remains stationary relative to ##m##?
 
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  • #83
43
2
On a side note, doesn't the equation ##\frac{dr}{dt}=-\frac{dx}{dt}## assume that ##M>>m## so that ##M## remains stationary relative to ##m##?
Yes, ##M## remains stationary relative to ##m##.
I would be happy to derive an equation for ##M## ~ ##m## so ##M## isn't stationary, but I don't have enough knolige in calculus.
 
  • #84
694
114
  • #85
19,802
4,048
I think I've found a neater solution to the problem. If you leave the equation in the form $$t=\int_{r_0}^r \frac{-dr}{\sqrt{\frac{2GM}{r} - \frac{2GM}{r_0}}}$$
and make the substitution ##r=\frac{r_0}{sec^2 \theta}##, you get
This is certainly not any "neater" than the substitution I recommended in post #34, ##r=r_0sin^2θ##, since, if ##r=\frac{r_0}{sec^2 \theta}##, then ##r=r_0cos^2θ##, which is essentially the same substitution. Your θ is just π/2 minus my θ. In posts #50, and 66-70, rcgldr employed my substitution, and obtained the same results you have obtained. Your parameter λ is exactly the same as your parameter θ. If you substitute ##π/2-θ## (Chet's θ) for λ in your final equation, you get rcgldr's results.

I was thinking about using these Taylor sequences to go further:
$$cos \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n}}{(2n)!}$$
$$sin \ \lambda = \sum_{n=0}^∞ \frac{(-1)^n \lambda^{2n+1}}{(2n+1)!}$$
However, this would then mean that the functions ##r(t)##, ##v(t)## and ##a(t)## would all have an infinite series, which would look very odd (pardon the parity pun). Is there any other way around this?
Using a Taylor series expansion is definitely not the way to solve this non-linear algebraic equation for r as a function of t. There are numerous methods available for numerical solution, including the half-interval technique and Newton's method. (Of course, you could say that Newton's method involves expanding to the linear term in a Taylor series, but this is not what I think you had in mind.)

If I wanted r as a function of t, I would just make a table of t(r) vs. r, and plot t as the abscissa and r as the ordinate.

Chet
 
  • #86
694
114
This is certainly not any "neater" than the substitution I recommended in post #34, ##r=r_0sin^2θ##, since, if ##r=\frac{r_0}{sec^2 \theta}##, then ##r=r_0cos^2θ##, which is essentially the same substitution. Your θ is just π/2 minus my θ. In posts #50, and 66-70, rcgldr employed my substitution, and obtained the same results you have obtained. Your parameter λ is exactly the same as your parameter θ. If you substitute ##π/2-θ## (Chet's θ) for λ in your final equation, you get rcgldr's results.


Using a Taylor series expansion is definitely not the way to solve this non-linear algebraic equation for r as a function of t. There are numerous methods available for numerical solution, including the half-interval technique and Newton's method. (Of course, you could say that Newton's method involves expanding to the linear term in a Taylor series, but this is not what I think you had in mind.)

If I wanted r as a function of t, I would just make a table of t(r) vs. r, and plot t as the abscissa and r as the ordinate.

Chet
OK Chet, I guess it was just me trying to avoid pi in the expression. I'd really appreciate if you can give an approximate expression for ##r(t)## though (or is the function implicit?).
 
  • #87
19,802
4,048
OK Chet, I guess it was just me trying to avoid pi in the expression. I'd really appreciate if you can give an approximate expression for ##r(t)## though (or is the function implicit?).
The function is implicit. So you really need to solve the non-linear algebraic equation for r(t).

Just make a table with column 1 being r and column 2 being t. Choose values of r in column 1 in equal increments (between the starting value r0 and the final value at the planet surface). Calculate the corresponding values of t from your equation, and fill them in in column 2. Then plot a graph of r vs t. This is the easiest thing to do. Just be happy that t can be expressed explicitly in terms of r.

Chet
 

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